Proof. Recall \(\varphi =\mathcal {F}\psi \)  and choose \(\varphi ^{\prime }\in \mathcal {C}^\infty _c(\mathbb {R}^3\setminus \lbrace 0\rbrace ,4)\)  with \(\parallel \!\!\varphi -\varphi ^{\prime }\!\!\parallel \le \varepsilon /2\) . Hence \(\parallel \!\!\psi -\psi ^{\prime }\!\!\parallel \le \varepsilon /2\)  for \(\psi ^{\prime }=\mathcal {F}^{-1}\varphi ^{\prime }\) . Choose \(0<v<\inf \lbrace \frac{|p|}{\epsilon (p)}:p\in \operatorname{supp}(\varphi ^{\prime })\rbrace \) . Let \(P_t := E^{\textsc {d}}(B_{v|t|})\) . Now, according to [1]}, there is a constant \(C_1\)  such that \(\parallel \!\!P_t\psi ^{\prime }_t\!\!\parallel \le C_1(1+|t|)^{-1}\)  for all \(t\) . Let \(\tau :=2C_1/\varepsilon \) . Then \(\parallel \!\!P_t\psi _t\!\!\parallel \le \parallel \!\!P_t(\psi _t-\psi ^{\prime }_t)\!\!\parallel  + \parallel \!\!P_t\psi ^{\prime }_t\!\!\parallel  \le \)  \(\parallel \!\!\psi -\psi ^{\prime }\!\!\parallel  +\, C_1(1+|t|)^{-1} \le \varepsilon \)  for \(|t|\ge \tau \) . — Now fix \(r>0\) . Then for \(|t|\ge \max \lbrace \tau , \frac{r}{v}\rbrace \)  one has \(\parallel \!\!E^{\textsc {d}}(B_r)\psi _t\!\!\parallel \le \parallel \!\!E^{\textsc {d}}(B_{v |t|})\psi _t\!\!\parallel \le \epsilon \) .
\(\Box \)
